1. 1 e – = electroplating: using electrolysis to deposit a thin layer of one metal onto another For electrolysis, current in amperes (1 A = 1 C/s) is passed through the liquid. One mole of transferred e – carries with it 96,500 C of charge. 1.60 x 10 –19 C Fe cathode to be plated with Ni Ni anode Ni 2+ (aq) V ext (Faraday’s constant) 6.02 x 10 23 e – =96,500 C ~

2. For how long must a 50.0 A current be passed through molten BaBr 2 in order to produce 500. g of barium? 500. g Ba = 14,100 s = 3.90 h Hall-Heroult process for purifying Al from Al 2 O 3 and Na 3 AlF 6 (1886) Paul Heroult (1863–1914) Charles Martin Hall (1863–1914) **To solve these problems, use unit cancellation.

3. For a voltaic cell, the maximum work the system can do on the surroundings is given by: w max =  G = –nFE where E is the reduction potential of the system at the conditions specified For an electrolytic cell, the work that the surroundings do (i.e., that the external “oomph” does) on the system is given by: w = nFE ext from external voltage source (E = E ext ) (i.e., it could be E o, but maybe not). w max WILL BE (–), and w WILL BE (+).

4. (1 h) The unit for electric power is the watt (1 W = 1 J/s). Electric companies often measure electrical energy in the kilowatt-hour (kWh). 1 kWh = 1000 W How many J is 1 kWh? = 3.6 x 10 6 J = 3.6 MJ To read an electric meter: 1. Read left to right. 2. Always round down. 8416

5. (Calc. w = nFE ext in Joules first, then convert to kWh.) Calculate the ideal number of kWh required to produce 10.0 kg of calcium from the electrolysis of molten calcium chloride if the applied emf is 75 V. Assume 100% efficiency. n = ?10,000 g Ca = 498.75 mol e – w = nFE ext = 498.75 (96,500)(75)= 3.61 x 10 9 J 3.61 x 10 9 J = 1.00 x 10 3 kWh In reality, electrolysis is only about 50% efficient, which means that half of the energy that is put in is given off as exhaust heat.